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(B) The Arrhenius equation is given by $k = A e^{-E_a / RT}$.Taking the natural logarithm,$\ln k = \ln A - \frac{E_a}{RT}$.Differentiating with respec

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$II$

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(B) The Arrhenius equation is given by $k = A e^{-E_a / RT}$.Taking the natural logarithm,$\ln k = \ln A - \frac{E_a}{RT}$.Differentiating with respect to $T$,$\frac{d(\ln k)}{dT} = \frac{E_a}{RT^2}$.For a small temperature change $\Delta T$,the change in $\ln k$ is $\Delta(\ln k) \approx \frac{E_a}{RT^2} \Delta T$.The percentage increase in rate constant is proportional to $\frac{\Delta k}{k} \approx \Delta(\ln k) = \frac{E_a \Delta T}{RT^2}$.To maximize the percentage increase,we need to maximize the value of $\frac{E_a}{T^2}$ (since $\Delta T$ is constant at $10 \ K$ for all cases).Comparing the cases:$I$: $\frac{40}{200^2} = \frac{40}{40000} = 1 	imes 10^{-3}$$II$: $\frac{80}{200^2} = \frac{80}{40000} = 2 	imes 10^{-3}$$III$: $\frac{40}{300^2} = \frac{40}{90000} \approx 0.44 	imes 10^{-3}$$IV$: $\frac{80}{300^2} = \frac{80}{90000} \approx 0.89 	imes 10^{-3}$The value is maximum for case $II$.

$Case$	$E_a \ (kcal/mol)$	$Temp. \ Change \ (K)$
$I$	$40$	$200 - 210$
$II$	$80$	$200 - 210$
$III$	$40$	$300 - 310$
$IV$	$80$	$300 - 310$

In which of the following cases is the percentage increase in rate constant maximum?
$Case$ $E_a \ (kcal/mol)$ $Temp. \ Change \ (K)$
$I$ $40$ $200 - 210$
$II$ $80$ $200 - 210$
$III$ $40$ $300 - 310$
$IV$ $80$ $300 - 310$

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In which of the following cases is the percentage increase in rate constant maximum?$Case$$E_a \ (kcal/mol)$$Temp. \ Change \ (K)$$I$$40$$200 - 210$$II$$80$$200 - 210$$III$$40$$300 - 310$$IV$$80$$300 - 310$